Class 10 Maths: Some Applications of Trigonometry – Extra Questions with Answers

Ans: A

Solution:
Let $h$ be the height of the tree.
We have $\tan(30^\circ) = \frac{h}{5}$.
Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have $\frac{1}{\sqrt{3}} = \frac{h}{5}$.
Thus, $h = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$.

Answer: $\frac{5\sqrt{3}}{3}$

Ans: C

Solution:
Let the length of the string be $s$. We can use the sine function.
$\sin(60^\circ) = \frac{10}{s}$
$s = \frac{10}{\sin(60^\circ)} = \frac{10}{\frac{\sqrt{3}}{2}} = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}$
Answer: $\frac{20\sqrt{3}}{3}$

Ans: D

Solution:
Let the cliff be AB, with A being the top. Let C and D be the positions of the two boats.
Angle of depression from A to C is $30^\circ$, and to D is $60^\circ$.
Thus, $\angle ACB = 30^\circ$ and $\angle ADB = 60^\circ$.
$AB = 80$ m.
In $\triangle ABD$, $\tan 60^\circ = \frac{AB}{BD} \implies \sqrt{3} = \frac{80}{BD} \implies BD = \frac{80}{\sqrt{3}}$.
In $\triangle ABC$, $\tan 30^\circ = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{80}{BC} \implies BC = 80\sqrt{3}$.
The distance between the two boats, $CD = BC – BD = 80\sqrt{3} – \frac{80}{\sqrt{3}} = 80\left(\sqrt{3} – \frac{1}{\sqrt{3}}\right) = 80\left(\frac{3-1}{\sqrt{3}}\right) = \frac{160}{\sqrt{3}} = \frac{160\sqrt{3}}{3}$.

Answer: $\frac{160\sqrt{3}}{3}$

Ans: C

Solution:
Since sin α = √2/2, α = 45° or π/4.
Since cos β = √3/2, β = 30° or π/6.
Therefore, α + β = 45° + 30° = 75° or π/4 + π/6 = 5π/12

Answer: 75°

Ans: A

Solution:
Let $L$ be the length of the ladder. The distance from the foot of the ladder to the wall is adjacent to the $60^\circ$ angle. We can use the cosine function to relate the adjacent side and the hypotenuse (the ladder). We have $\cos(60^\circ) = \frac{5}{L}$. Since $\cos(60^\circ) = \frac{1}{2}$, we have $\frac{1}{2} = \frac{5}{L}$. Solving for $L$, we get $L = 10$ meters.
Answer: 10 meters

Ans: A

Solution: Let $h$ be the height of the tower and $w$ be the width of the river.
From the first observation, $\tan(45^\circ) = \frac{h}{w}$, so $h = w$.
From the second observation, $\tan(30^\circ) = \frac{h}{w+10}$, so $\frac{1}{\sqrt{3}} = \frac{h}{w+10}$.
Substituting $h=w$, we get $\frac{1}{\sqrt{3}} = \frac{w}{w+10}$, so $w+10 = w\sqrt{3}$.
Then $w(\sqrt{3}-1) = 10$, and $w = \frac{10}{\sqrt{3}-1} = \frac{10(\sqrt{3}+1)}{2} = 5(\sqrt{3}+1)$.
Since $h=w$, $h = 5(\sqrt{3}+1)$.
Therefore, the height of the tower is $5(1+\sqrt{3})$ meters and the width of the river is $5(1+\sqrt{3})$ meters.

Answer: $h = 5(1+\sqrt{3})$ meters, $w = 5(1+\sqrt{3})$ meters

Ans: B

Solution:
Let $h$ be the height of the pole, and $x$ be the initial distance from the point on the ground to the base of the pole.
From the first observation, we have $\tan 45^\circ = \frac{h}{x}$, so $1 = \frac{h}{x}$, which means $x = h$.
From the second observation, we have $\tan 60^\circ = \frac{h}{x-10}$, so $\sqrt{3} = \frac{h}{x-10}$.
Substituting $x = h$, we have $\sqrt{3} = \frac{h}{h-10}$.
Thus, $h\sqrt{3} – 10\sqrt{3} = h$, so $h(\sqrt{3}-1) = 10\sqrt{3}$.
Therefore, $h = \frac{10\sqrt{3}}{\sqrt{3}-1} = \frac{10\sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{10(3+\sqrt{3})}{2} = 5(3+\sqrt{3})$.

Answer: $5(3+\sqrt{3})$

Ans: B

Solution:
Let the height of the tower be $h$. Let the distance from the first point to the base of the tower be $x$, and the distance from the second point to the base of the tower be $x-1$.
From the first point, $\tan 45^\circ = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x=h$.
From the second point, $\tan 60^\circ = \frac{h}{x-1} \Rightarrow \sqrt{3} = \frac{h}{x-1} \Rightarrow x-1 = \frac{h}{\sqrt{3}}$.
Since $x=h$, we have $h-1 = \frac{h}{\sqrt{3}} \Rightarrow h – \frac{h}{\sqrt{3}} = 1 \Rightarrow h(1 – \frac{1}{\sqrt{3}}) = 1 \Rightarrow h = \frac{1}{1-\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{\sqrt{3}-1} = \frac{\sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{3+\sqrt{3}}{2}$.
Answer: $\frac{3+\sqrt{3}}{2}$

Ans: A

Solution:
Let $h$ be the height of the flagpole. Let $d$ be the distance from the point on the ground to the building.
From the given information, we have:
$\tan(60^\circ) = \frac{30}{d} \implies d = \frac{30}{\tan(60^\circ)} = \frac{30}{\sqrt{3}} = 10\sqrt{3}$
$\tan(75^\circ) = \frac{30+h}{d} \implies 30+h = d \tan(75^\circ)$
We know that $\tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan(45^\circ)+\tan(30^\circ)}{1-\tan(45^\circ)\tan(30^\circ)} = \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{3+2\sqrt{3}+1}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.
So, $30+h = 10\sqrt{3}(2+\sqrt{3}) = 20\sqrt{3} + 30$
Thus, $h = 20\sqrt{3}$.

Answer: $20\sqrt{3}$

Ans: D

Solution:
Let the height of the lighthouse be $h$.
Let the distance from the lighthouse to point A be $x$.
Then the distance from the lighthouse to point B is $60 – x$.
From point A, $\tan(60^\circ) = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$
From point B, $\tan(45^\circ) = \frac{h}{60-x} \Rightarrow 60 – x = h$
Substituting $x = \frac{h}{\sqrt{3}}$ into $60 – x = h$, we get $60 – \frac{h}{\sqrt{3}} = h$
$60 = h + \frac{h}{\sqrt{3}}$
$60 = h(1 + \frac{1}{\sqrt{3}})$
$h = \frac{60}{1 + \frac{1}{\sqrt{3}}} = \frac{60\sqrt{3}}{\sqrt{3} + 1} = \frac{60\sqrt{3}(\sqrt{3}-1)}{3-1} = \frac{60(3-\sqrt{3})}{2} = 30(3-\sqrt{3})$

Answer: $30(3-\sqrt{3})$